pandas.Series.str.partition — pandas 0.24.0rc1 documentation (original) (raw)
Series.str. partition(sep=' ', expand=True)[source]¶
Split the string at the first occurrence of sep.
This method splits the string at the first occurrence of sep, and returns 3 elements containing the part before the separator, the separator itself, and the part after the separator. If the separator is not found, return 3 elements containing the string itself, followed by two empty strings.
| Parameters: | sep : str, default whitespace String to split on. pat : str, default whitespace Deprecated since version 0.24.0: Use sep instead expand : bool, default True If True, return DataFrame/MultiIndex expanding dimensionality. If False, return Series/Index. |
|---|---|
| Returns: | DataFrame/MultiIndex or Series/Index of objects |
Examples
s = pd.Series(['Linda van der Berg', 'George Pitt-Rivers']) s 0 Linda van der Berg 1 George Pitt-Rivers dtype: object
s.str.partition() 0 1 2 0 Linda van der Berg 1 George Pitt-Rivers
To partition by the last space instead of the first one:
s.str.rpartition() 0 1 2 0 Linda van der Berg 1 George Pitt-Rivers
To partition by something different than a space:
s.str.partition('-') 0 1 2 0 Linda van der Berg 1 George Pitt - Rivers
To return a Series containining tuples instead of a DataFrame:
s.str.partition('-', expand=False) 0 (Linda van der Berg, , ) 1 (George Pitt, -, Rivers) dtype: object
Also available on indices:
idx = pd.Index(['X 123', 'Y 999']) idx Index(['X 123', 'Y 999'], dtype='object')
Which will create a MultiIndex:
idx.str.partition() MultiIndex(levels=[['X', 'Y'], [' '], ['123', '999']], labels=[[0, 1], [0, 0], [0, 1]])
Or an index with tuples with expand=False:
idx.str.partition(expand=False) Index([('X', ' ', '123'), ('Y', ' ', '999')], dtype='object')